Find the value of $\theta$ This is a question from a physics book that I can't get my head around.
While moving in, a new homeowner is pushing a box across the floor at a constant velocity. The coefficient of kinetic friction between the box and the floor is 0.41. The pushing force $F$ is directed downwards at an angle $\theta$ below the horizontal. When $\theta$ is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. Find the value of $\theta$.
Horizontal: $$F\cos\theta = Fr = \mu_{k}R\tag{1}$$
Vertical: $$R = mg+F\sin\theta, \tag{2}$$
Together: $$F\cos\theta = \mu_{k}(mg+F\sin\theta),\tag{3}$$
The answer seems to go from: $$F\cos\theta=\mu_{k}F\sin\theta$$
Which for: $\mu_{k}=0.41$, gives: $\theta = 67.7deg$
But where did the $mg$ in Eq. 3 go ?
 A: Lets look at this problem in the context of large applied forces.
We can push the box as long as $Fcos(\theta) > \mu R$. Slightly rewriting your third equation this means the box moves as long as
$$cos(\theta) > \mu (\frac{mg}{F} + sin(\theta))$$
Lets say we push at an angle such that $cos(\theta) = \mu sin(\theta) + \delta$ with $\delta > 0$. If this is the case we can move the box as long as we push with a force F that satisfies
 $$\frac{\mu mg}{F} < \delta  $$ 
No matter how small $\delta$ is we can always push hard enough to move the box. The only value of $\delta$ for which no F can satisfy this equation is if $\delta = 0$. So, if we are looking for an angle at which no finite force can move the box we need to solve
$$cos(\theta) = \mu sin(\theta)$$ 
A: The analysis already provided by Zarathustra is clever and correct, but since your question was what happened to mg in equation (3), the direct answer would be that no matter how large m is, F can always be made large enough to make mg negligible.    Then, when mg is ignored, F divides out.
A: Ignore the book and find the missing mg yourself. Immediately draw a picture of your free body diagram (always always always DO THIS!) which I will describe for you as you draw it. There should be a box representing the homeowners junk of mass $m$. From this box, there should be a vector for each force acting on the box. This will include one straight down with the tag $mg$ which was likely one of the first things you were drawn in class. Next there should be a vector indicating the force of the homeowner which we can call $\vec{F_{}}$ having magnitude $F$ and should be pointing to (I chose) the right and downward labeling the angle between it and your x-axis, the angle $\theta$. Are we done? No, We haven't labeled the friction since we know it's present. There will be a frictional force proportional to the normal force, ah! lets write the normal force first. The normal force, typically indicated as $\vec{F}_N$ will always point... you know where... perpendicular to the surface. It should have a magnitude (which you will write to the side as not to clutter your pretty picture) of... well as it seems... Some component of the homeowners force + the $mg$ that points opposite to it. It seems $\vec{F}_N = Fsin(\theta)+mg$ and is pointing straight up. Now we write down friction since we know $\vec{F}_N$, so the friction will always oppose the applied force and be along the axis of the motion (here the $x$ axis) and so I have mine pointing to the left. You should have something to the side that says that $F_f=\mu F_N$ There! all done.
Now add up all the x direction forces you have written down and repeat for the y direction separately. Each will be equal to the acceleration in the respective direction, and since at current there is no acceleration in the y, your y forces should sum to zero. Now for the x-direction (which answers your question), you should get something like $Fcos(\theta) - F\mu sin(\theta) -mg \mu = ma_x $ but we want to know how the $a_x$ could be zero so let's set it equal to zero and see what this gives us. Should give something like this
$$cos(\theta) - \mu sin(\theta) = \frac{mg \mu}{F}    $$
where I've put all the angle content on one side because that's what we're interested in. Now if the homeowner were to push infinitely hard, the right side would go to zero. So if we now set that side equal to zero, we are looking at the condition for 1-no acceleration in the $x$ and 2-an infinitely strong homeowner. well it seems the only way that could happen is if $cos(\theta) = \mu sin(\theta)$ which looks a whole lot like what you've presented in equation (not labeled). By dividing both sides by $sin(\theta)$, you'll see that gives the condition $cot(\theta) = \mu$ and so to find $\theta$ you take the inverse tangent of $\frac{1}{\mu}$
